Projectile Motion - Finding initial velocity

Question

You need to send a bowling ball up exactly $205$ meters so someone at the top of the Washington Monument can take a picture of it "hanging" in space. The bowling ball will be shot from a mortar tube from a platform $25$ meters above the ground. You can determine what the initial velocity is because the technician will put in enough black powder to give the bowling ball whatever initial velocity you tell him to give you. So....

(1) What initial velocity ($v_o$) will cause the bowling ball to "freeze" in mid-air at exactly $205$ meters above the ground?

(2) How fast will the bowling ball be moving right before it hits the ground?

Using The formula $s(t) = s_0 + v_0(t) + .5gt^2$

$s_0 = 25$ and $t = 18$ sec and $g = 10$ m/sec$^2$

Answer 1

Use conservation of energy.

At 25 m above the ground, energy = kinetic + potential. Kinetic = $(1/2) m v_0^2$. Potential = $m g h$. $m$ = mass of ball, $g$ = acceleration due to gravity, $h$ = 25 m.

At $H$ = 205 m above the ground, the ball is changing direction, so the velocity is...? And therefore the kinetic energy is...? The potential energy is $m g H$. Now set the total energy at 25 m and 205 m equal to each other. You should have only one unknown after cancellation.

At the ground, the potential energy is...?

Answer 2

You need the ball to rise $180$ meters, so it should start at the speed it would have if it fell $180$ meters. Do you know formulas for the velocity and distance covered under gravity?

Answer 3

This is probably a better fit for Physics.SE...

I'm going to assume you are ignoring air resistance, as there's no Differential Equations tag.

Part 1

We have from conservation of energy: $$\Delta K + \Delta U_g = 0$$ $$\frac{1}{2}m\left(v_f^2-v_0^2\right) + mg(y_f-y_0) = 0$$ Dividing by $m$ and plugging in values: $$\frac{1}{2}\left(-v_0^2\right) + g(205\text{m}-25\text{m}) = 0$$ Simplify and solve for $v_0$: $$\frac{1}{2}\left(v_0^2\right) = g(205\text{m}-25\text{m})$$ $$v_0 = \sqrt{2g(205\text{m}-25\text{m})}$$ $$v_0 = \sqrt{2(9.80)\text{m/s$^2$}(180\text{m})}$$ $$v_0 = \sqrt{3528}\frac{\text{m}}{\text{s}}\approx 59.39\frac{\text{m}}{\text{s}}$$

Part 2

The ball will fall $205$ meters. Again, conservation of energy: $$\Delta K + \Delta U_g = 0$$ $$\frac{1}{2}m\left(v_f^2-v_0^2\right) + mg(y_f-y_0) = 0$$ Dividing by $m$ and plugging in values: $$\frac{1}{2}\left(v_f^2\right) + g(-205\text{m}) = 0$$ $$v_f^2 = 2g(205)\text{m}$$ $$v_f = \sqrt{9.80\frac{\text{m}}{\text{s}^2}(410)\text{m}}$$ $$v_f \approx 63.39\frac{\text{m}}{\text{s}}$$

With Kinematic Equations

I guess this is supposed to be solved with the kinematic equations: $$s = s_0 + v_0t - \frac{gt^2}{2}$$ $$v(t) = v_0 - gt$$ (With $g$ being the magnitude of gravity ($9.80\frac{\text{m}}{\text{s}^2}$))

The first step is to solve for the time it takes to reach the top. The easiest way is to use the equation: $$v(t) = v_0 - gt$$ Thus, we have: $$0 = v_0 - gt_{top}$$ $$t_{top} = \frac{v_0}{g}$$ Now we plug this in to the other equation: $$s = s_0 + v_0t - \frac{gt^2}{2}$$ $$s_{top} = s_0 + v_0\left(\frac{v_0}{g}\right) - \frac{g\left(\frac{v_0}{g}\right)^2}{2}$$ $$205 = 25 + \frac{v_0^2}{g} - \frac{v_0^2}{2g}$$ $$180 = \frac{2v_0^2-v_0^2}{2g}$$ $$180 = \frac{v_0^2}{2g}$$ $$v_0^2 = 180(2g)$$ $$v_0 = \sqrt{360\cdot9.80\frac{\text{m}^2}{\text{s}^2}} \approx 59.4 \frac{\text{m}}{\text{s}}$$

The falling speed is very similar, so I will not work it out for you.